Question

Find the equation of all lines having slope $0$which are tangent to the curve $y=\frac{1}{x^2-2x+3}$

Answer 1

Given curve is $y=\frac{1}{x^2-2x+3}\cdots \left(i\right)$
Slope of tangent is $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{x^2-2x+3}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{d(x^2-2x+3{)}^{-1}}{dx}$
$\Rightarrow \frac{dy}{dx}=-1(x^2-2x+3{)}^{-2}.\frac{d}{dx}(x^2-2x+3)$
$\Rightarrow \frac{dy}{dx}=-(x^2-2x+3{)}^{-2}(2x-2)$
$\Rightarrow \frac{dy}{dx}=-2(x^2-2x+3{)}^{-2}(x-1)$
$\Rightarrow \frac{dy}{dx}=\frac{-2(x-1)}{(x^2-2x+3{)}^2}$
Given slope of tangent is $0$
$\Rightarrow \frac{dy}{dx}=0$
$\Rightarrow \frac{-2(x-1)}{(x^2-2x+3{)}^2}=0$
$\Rightarrow x=1$
Putting $x=1$ in equation $\left(i\right):$
$y=\frac{1}{x^2+2x+3}$
$\Rightarrow y=\frac{1}{(1{)}^2-2(1)+3}$
$\Rightarrow y=\frac{1}{2}$
Point is $(1,\frac{1}{2})$
Now, equation of tangent passing through point $(1,\frac{1}{2})$ and slope $0$, is $(y-\frac{1}{2})=0(x-1)$
$\Rightarrow y-\frac{1}{2}=0$
$\Rightarrow y=\frac{1}{2}$
Hence, the equation of tangent is $y=\frac{1}{2}.$